Hello, Bemcma13!

You have a circular disk and you removed a wedge from that disk.

Then reconnected to two edges where the wedge had been removed

to create a cone, positioned that cone on a flat surface and measured

the angle between the flat surface and the the wall of the cone to the vetex.

What relationship would that angle have with the angle of the wedge

which was initially removed?

We have a circular disk of radius $\displaystyle R$.

We remove a sector with central angle $\displaystyle \theta.$

Code:

* * *
* *
* *
* *
* *
* * *
* /@\ *
R /:::\ R
* /:::::\ *
* /:::::::\ *
* :::::::::*
* * *

The length of the major arc is: $\displaystyle (2\pi-\theta)R$

This is the circumference of the circular base of the cone.

The side view of the cone looks like this:

Code:

*
/|\
/ | \
/ | \
R / | \ R
/ | \
/ | \
/ | α \ π-α
*-------+-------*------
: - r - :

The slant height is $\displaystyle R.$

The radius of the base is $\displaystyle r.$

The base angle is $\displaystyle \alpha.$

You seek its supplement: $\displaystyle \pi-\alpha$

The circumference of the base circle is: $\displaystyle (2\pi - \theta)R$

We have: .$\displaystyle 2\pi r \:=\:(2\pi-\theta)R \quad\Rightarrow\quad r \:=\:\frac{(2\pi-\theta)R}{2\pi}$

. . Then: .$\displaystyle \cos\alpha \:=\:\frac{r}{R} \:=\:\frac{2\pi-\theta}{2\pi} \:=\:1 - \frac{\theta}{2\pi}$

. . Hence: .$\displaystyle \alpha \:=\:\cos^{-1}\!\left(1 - \tfrac{\theta}{2\pi}\right) $

Therefore: .$\displaystyle \pi - \alpha \;=\;\pi - \cos^{-1}\!\left(1 -\tfrac{\theta}{2\pi}\right)$